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Journey of The Dog

Jessica, Warner decided to meet & left their home. And a puppy starts walking down a road. They started at the same time. Their homes are located at 33 KM away from each other.

• Warner walks at 5 miles/hour.
• Jessica walks at 6 miles/hour.
  
The puppy runs from Warner to Jessica and back again with a constant speed of 10 miles/hour.
 
The puppy does not slow down on the turn. How far does the puppy travel in till Jessica and Warner meet?


Distance covered by the dog in the journey?

Know here the distance traveled by puppy! 

Distance in The Dog's Journey!


What was the puzzle? 

First thing on which we need to focus on in how much time Jessica and Warner would meet. Since they are moving towards each other the distance of 33 KM is being covered at 5 + 6 = 11 KM/h.  So they are going to meet each other in 33/11 = 3 hours. Now everything else here can deceive you to find distance covered by puppy. All you need to do is stick to the basics.

Speed = Distance / Time

Distance = Speed * Time 

Distance covered by Puppy = Speed of Puppy * Time for which it traveled.

Distance covered by Puppy = 10 * 3 = 30 KM 

Calculation of the distance covered by the dog in the journey

So Puppy travels 30 KM to & fro until Jessica and Warner meets. 

 

Equate Number of Heads or Tails

You are blindfolded and 10 coins are placed in front of you on the table. You are allowed to touch the coins but can't tell which way up they are by feel. You are told that there are 5 coins head up, and 5 coins tail up but not which ones are which.

How do you make two piles of coins each with the same number of heads up? You can flip the coins any number of times.

Equate number of heads/tails in 2 piles


This is how it can be done! 

Trick To Equate Number of Heads or Tails


What was the task? 

Without thinking too much we need to make 2 piles of 5 coins each. Now there are 3 possibilities here depending on number of heads in either pile. One of the pile might have either 0 or 1 or 2 heads (other having 5 or 4 or 3 heads).

Case 1 : 

P1 : T T T T T
P2 : H H H H H

Case 2 : 

P1 : H T T T T
P2 : H H H H T

Case 3 :

P1 : H H T T T
P2 : H H H T T

Now just flipping all the coins from single pile will make number of heads (or say tails) in both piles equal. So we can flip coins of either P1 or P2. Let's flip all coins of P2.


Case 1 : 

P1 : T T T T T         Number of heads - 0
P2 : T T T T T         Number of heads - 0

Case 2 : 

P1 : H T T T T         Number of heads - 1
P2 : T T T T H         Number of heads - 1

Case 3 :

P1 : H H T T T         Number of heads - 2
P2 : T T T H H         Number of heads - 2

Flipping Coins of 1 Pile To Equate Number of Heads/Tails

Challenge of Grouping The Coins

You are given a unlimited number of coins and 10 pouches. Now, you have to divide these coins in the given pouches in a manner that if someone asks you for any number of coins between 1 to 1000, you should be able to give the amount by just giving the pouches. You are not allowed to open pouches for that.

How will you do it? 



Make 10 pouches of coins to produce any number from 1 to 1000.


Know here the only efficient way to do that! 

Source 

Grouping The Coins in Binary Numbers


What was the challenge? 

Once again here binary number system comes in handy. Similar kind of use of binary system in day to day life is here! Another intelligent use is here!  We are already provided 10 pouches which is exactly equal to the number of bits required to represent any number from 1 to 1000. Let's number the pouch as Pouch 0 to Pouch 9. So we need to group coins in 10 pouches like below.

Pouch 0 : 1
Pouch 1 : 2
Pouch 2 : 4
Pouch 3 : 8
Pouch 4 : 16
Pouch 5 : 32
Pouch 6 : 64
Pouch 7 : 128
Pouch 8 : 256
Pouch 9 : 512


Now if somebody asks us for 30 coins then we should give Pouch 4, Pouch 3, Pouch 2, Pouch 1. (11110) That's the binary representation of 30 if we assume Pouches as a bits. If another asks for 828 (binary - 1100111100) then we should give Pouch 9, Pouch 8, Pouch 5, Pouch 4, Pouch 3,
Pouch 2.     

Using binary system in grouping the coins!

Identify The Cards

From a pack of 52 cards ,I placed 4 cards on the table.

I will give you 4 clues about the cards:


Clue 1: Card on left cannot be greater than card on the right.
Clue 2: Difference between 1st card and 3rd card is 8.
Clue 3: There is no card of ace.
Clue 4: There is no face cards (queen,king,jacks).
Clue 5: Difference between 2nd card and 4th card is 7.


Identify four cards ?


Identify The Cards using clues given

Cards identified here!

Source 

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